Question

Figure 5-55 gives, as a function of time t, the force component x that acts on a 3.40 kg ice block that can move only along the x axis. At t = 0, the block is moving in thepositive direction of the axis, with a speed of 3.3 m/s. What is its velocity, including sign, at t = 11 s?

Answer 1

The acceleration \(a=\frac{1}{m}\) is the derivative of the

$$ \begin{array}{l} \text { velocity. As } a=\frac{d v}{d t} \\ d v=a d t=\frac{F}{m} d t \end{array} $$

Hence the velocity is the time inegral of \(\frac{\mathrm{F}}{\mathrm{m}}\)

The area of the given graph gives F.dt Hence \(A=F \cdot d t\)

Now the area can be determined by dividing the

graph into triangles and rectangles.

$$ \begin{array}{l} \begin{aligned} \mathrm{A} &=\mathrm{F} \cdot \mathrm{dt}=\operatorname{ar}(A B M O)+\operatorname{ar}(\square \beta C M N)+\operatorname{ar} \Delta N D C+\operatorname{ar}(\operatorname{trap} z D G P E) \\ &=\frac{(2+6)}{2} \times 2+6 \times 4+\frac{1}{2} \times 1 \times 6+\frac{1}{2}(2+5)(-4) \\ &=8+18+3-14 \\ \Rightarrow & \mathrm{F} \cdot \mathrm{dt}=15 \mathrm{~N} \mathrm{~s} \\ \text { Hence } \mathrm{v}-\mathrm{v}_{0} &=\frac{15}{\mathrm{~m}} \end{aligned} \end{array} $$

Given \(m=3.4 \mathrm{~kg}\)

and \(\mathrm{v}_{0}=3.3 \mathrm{~m} / \mathrm{s}\)

$$ \begin{array}{l} \text { Hence } v=\frac{15}{m}+3.3=\frac{15}{3.4}+3.3 \\ \Rightarrow v=4.411+3.3=+7.711 \mathrm{~m} / \mathrm{s} \end{array} $$

or \(\mathrm{v} \approx+7.7 \mathrm{~m} / \mathrm{s}\)

The positive value of velocity indicates that

\(\mathrm{v}\) is in the \(+\mathrm{x}\) direction.